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Solve this system of linear equations:

[tex]\[
4x - 2y = 8 \quad y = \frac{3}{2}x - 2
\][/tex]

Steps:
1. Plot the [tex]\( x \)[/tex]-intercept of the first equation.
2. Plot the [tex]\( y \)[/tex]-intercept of the first equation.
3. Plot the [tex]\( y \)[/tex]-intercept of the second equation.
4. Use the slope of the second equation to plot one more point.

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & -4 \\
0 & -2 \\
1 & 0 \\
2 & 0 \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To solve the given system of linear equations:

[tex]\[ 4x - 2y = 8 \quad \text{and} \quad y = \frac{3}{2}x - 2 \][/tex]

we can follow the steps provided.

### Step 1: Plot the [tex]\( x \)[/tex]-intercept of the first equation.

To find the [tex]\( x \)[/tex]-intercept of the equation [tex]\( 4x - 2y = 8 \)[/tex], set [tex]\( y = 0 \)[/tex]:

[tex]\[ 4x - 2(0) = 8 \\ 4x = 8 \\ x = 2 \][/tex]

So the [tex]\( x \)[/tex]-intercept is at the point [tex]\( (2, 0) \)[/tex].

### Step 2: Plot the [tex]\( y \)[/tex]-intercept of the first equation.

To find the [tex]\( y \)[/tex]-intercept of the first equation, set [tex]\( x = 0 \)[/tex]:

[tex]\[ 4(0) - 2y = 8 \\ -2y = 8 \\ y = -4 \][/tex]

So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -4) \)[/tex].

### Step 3: Plot the [tex]\( y \)[/tex]-intercept of the second equation.

For the second equation [tex]\( y = \frac{3}{2}x - 2 \)[/tex], set [tex]\( x = 0 \)[/tex]:

[tex]\[ y = \frac{3}{2}(0) - 2 \\ y = -2 \][/tex]

So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -2) \)[/tex].

### Step 4: Use the slope of the second equation to plot one more point.

The slope of the second equation, [tex]\( y = \frac{3}{2}x - 2 \)[/tex], is [tex]\( \frac{3}{2} \)[/tex]. This means for every increase in [tex]\( x \)[/tex] by 1, [tex]\( y \)[/tex] increases by [tex]\( \frac{3}{2} \)[/tex].

Starting from the [tex]\( y \)[/tex]-intercept [tex]\( (0, -2) \)[/tex], if [tex]\( x \)[/tex] increases by 1:

[tex]\[ x = 1 \\ y = \frac{3}{2}(1) - 2 \\ y = \frac{3}{2} - 2 \\ y = -0.5 \][/tex]

So, another point is [tex]\( (1, -0.5) \)[/tex].

### Conclusion:

From the above steps, we have identified the following points:

- [tex]\( (0, -4) \)[/tex] from the first equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (0, -2) \)[/tex] from the second equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (2, 0) \)[/tex] from the first equation's [tex]\( x \)[/tex]-intercept.
- [tex]\( (1, -0.5) \)[/tex] from the slope of the second equation.

Putting these points into the table, we get:

[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -4 \\ 0 & -2 \\ 2 & 0 \\ 1 & -0.5 \\ \hline \end{array} \][/tex]

So, we have plotted the intercepts and a few other key points which would help in graphically solving or verifying the system of equations.
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